Problem 3 - Calculus for Biology and Medicine

Question

Let \(f(x)=\sqrt{x}+x, \quad 1 \leq x \leq 2\). (a) Graph \(y=f(x)\) for \(1 \leq x \leq 2\). (b) Use the Intermediate Value Theorem to conclude that \(\sqrt{x}+x=3\) has a solution in \((1,2)\).

Step-by-Step Solution

Answer

The equation \( \sqrt{x} + x = 3 \) has a solution in \((1, 2)\) by the Intermediate Value Theorem.

Step 1: Understand the Function

The function given is \( f(x) = \sqrt{x} + x \). We are working within the closed interval \([1, 2]\). This function consists of two parts: the square root function \( \sqrt{x} \) and the linear function \( x \).

Step 2: Graph the Function

To graph \( f(x) = \sqrt{x} + x \) for \( 1 \leq x \leq 2 \), first, evaluate the function at key points. Calculate \( f(1) = \sqrt{1} + 1 = 2 \) and \( f(2) = \sqrt{2} + 2 \approx 3.414 \). Plot the points \((1, 2)\) and \((2, 3.414)\) and sketch the curve that connects them. The curve will be smooth and slightly concave upward as \( \sqrt{x} \) increases slower than linear \(x\).

Step 3: Identify the Equation within the IVT Context

We need to show that the equation \( \sqrt{x} + x = 3 \) has a solution in the interval \((1, 2)\). This translates to finding \( x \) such that \( f(x) = 3 \). According to the graph, \( f(1) = 2 \) and \( f(2) \approx 3.414 \), indicating there is a value of \( x \) in \((1, 2)\) for which \( f(x) = 3 \).

Step 4: Apply the Intermediate Value Theorem

The Intermediate Value Theorem (IVT) states that if a function \( f(x) \) is continuous on \([a, b]\) and \( N \) is any number between \( f(a) \) and \( f(b) \), then there exists at least one \( c \) in \((a, b)\) such that \( f(c) = N \). Here, \( f(x) \) is continuous (since both \( \sqrt{x} \) and \( x \) are continuous), \( f(1) = 2 \), \( f(2) \approx 3.414 \), and 3 is between 2 and 3.414. By IVT, there is at least one value \( c \) in \((1, 2)\) where \( f(c) = 3 \).

Step 5: Conclude the Existence of a Solution

The Intermediate Value Theorem confirms that since \( f(x) \) is continuous and 3 is between \( f(1) \) and \( f(2) \), there is a solution to \( \sqrt{x} + x = 3 \) in the interval \((1, 2)\). Thus, there exists at least one \( x \) in \((1, 2)\) where \( \sqrt{x} + x = 3 \).

Key Concepts

Continuous Function

In the world of mathematics, a continuous function is one that has no breaks, jumps, or holes over its domain. Simply put, you can draw it without lifting your pencil from the paper.
For the function \( f(x) = \sqrt{x} + x \), both \( \sqrt{x} \) and \( x \) are continuous over the interval \([1, 2]\). This is essential for using the Intermediate Value Theorem to find solutions.

• The function \( \sqrt{x} \) is continuous because as \( x \) increases, \( \sqrt{x} \) gently increases without interruption.
• The linear function \( x \) is also continuous, as it increases steadily and linearly without any disruptions.

Knowing that \( f(x) \) is continuous ensures that it smoothly transitions from one function value to another, allowing us to apply important mathematical theorems like the Intermediate Value Theorem.

Interval Analysis

Interval analysis involves examining the behavior of a function within a specific range or interval. For \( f(x) = \sqrt{x} + x \), we're focusing on the interval \([1, 2]\).

• Starting at \( x = 1 \), we find \( f(1) = \sqrt{1} + 1 = 2 \).
• At the other endpoint, \( x = 2 \), we compute \( f(2) = \sqrt{2} + 2 \approx 3.414 \).

This shows that within the interval \([1, 2]\), \( f(x) \) spans from 2 to approximately 3.414. Interval analysis helps determine the range of possible function values, crucial when applying the Intermediate Value Theorem. It tells us that since 3 falls between 2 and 3.414, there must be some \( x \) in \((1, 2)\) such that \( f(x) = 3 \).

Graphing Functions

Graphing functions is a visual way to understand their behavior. For a function like \( f(x) = \sqrt{x} + x \), graphing offers a clear picture of how the function behaves over the interval \([1, 2]\).
To draw the graph:

• Plot the points \((1, 2)\) and \((2, 3.414)\), which are calculated values of the function at the bounds of the interval.
• Sketch a smooth curve connecting these points. It will gently rise due to the nature of \( \sqrt{x} \) being slowly increasing, combined with the steady slope from \( x \).

The graph not only confirms continuity visually but also allows us to see that any horizontal line between the values 2 and 3.414 will intersect the curve, emphasizing the existence of solutions within the interval.

Existence of Solutions

The concept of the existence of solutions often involves verifying that a particular function value can be achieved within a specific interval. The Intermediate Value Theorem is key to this verification for any continuous function.
For \( f(x) = \sqrt{x} + x \):

• We've identified that \( f(x) \) is continuous over the interval \([1, 2]\).
• We know that \( f(1) = 2 \) and \( f(2) \approx 3.414 \).
• We are interested in the value 3, which lies between 2 and 3.414.

By the Intermediate Value Theorem, since 3 is between the values that \( f(x) \) takes at the ends of the interval, there exists some \( c \) in \( (1, 2) \) where \( f(c) = 3 \). This guarantees that there is a solution to the equation \( \sqrt{x} + x = 3 \) within this interval.